Avalanche breakdown (I think) despite not violating my BJT's maximums (I think)

I bought a kit of various values of BJT for my project from Amazon. I'm suspecting they're not up to spec because they're cheap but I'm not sure. I'm trying to use BC327.

Here's my circuit

It works perfectly when connected to V1=5V but Q7 leaks a lot of current (hundreds of milliamps) when connected to my desired V1=16V. I looked this up and a search suggested it's breakdown because of excessive voltage, but the spec sheet says it's good to 45 volts.

Here's what the simulation is saying that transistor is experiencing, which all seems reasonable.

image

Is this a situation where I should buy new, higher quality BC327 and it'll probably work, or am I not understanding the problem?

Thanks for your help!

A quick look; Q8 should be fully saturated with 50mA base current. That would mean R5 has close to (16V-0.7V)/1k = 15.4mA and I would expect most of this to flow through Q7's base. Your modelling shows 1.3mA. A big difference.
Can you resolve this disparity?

I wouldn't.

As Q7 turns on more current flows through R2, lowering the voltage across Q7 BE, diverting the R5 current through D4 D5. At some point Q7 becomes a constant current source.

All true. Perhaps I'm mistaken (often :wink: ) . If D5 and Q7's Vbe are comparable, then the volt drop across R2 would be comparable to the drop across D4; at ~0.65V (say) the curren tthrough R2 would be ~1.5A. What am I missing?

I wonder how Q8 is supposed to draw 50 mA base current, from a 16V source, through a 1K resistor.

I simulated this simplified circuit, and got 1A through R3, with the BC327 dissipating 13W (over 20x the rated power dissipation). It is not a viable design.

Capture

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Q8 will be fully on so may as well be a short circuit with R5 connected to 0V. I got the current through R2 at over 2A, I could not see my mistake so I left it for someone else to reply.

Why are you surprised Q7 is conducting?
Where did you find mOhm resistors... not in your standard kit I expect...

I just saw the constant current a took it at face value, I did add the caveat

That doesn't look like a constant current to me.

The benefit of a simulation :slight_smile:

It is. That is how that standard circuit works. Presuming that the transistor survives the abuse.

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The current flow through R3 doesn't change with the base current of Q8. Pretend it's 1 mA, it simulates the same.

It's above the maximum power of Q8 in the worst-case scenario (variable resistance) but it needs to be on for literally less than one second a day. I think it'll survive but if it doesn't I'll work on it.

I'm curious why your simulation doesn't match mine, which gives only half an amp through R3. Edit: they don't match because I have the wrong resistor in R1. I was tooling around with it. It should be 1.2 ohms.

Two possibilities come to mind.

  1. We aren't simulating the same circuit.
  2. We aren't using the same simulators.

Have fun with your project!

I edited. I had the wrong resistor in R1. It should be 1.2 ohms for 500 mA.

That is constant over time!
What about varying load?

If you are not familiar with that circuit, simulate it with different loads and see how it behaves. It is quite useful as an approximate constant current source.

Or breadboard it. The more realistic example below should maintain Ic at about 0.65V/1K or 650 uA. The LTSpice simulation predicts 637 uA though R3.

Capture

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The resistance varies between 2.5 and 30 ohms. With the 1.2 ohm resistor R2 (doesn't match the circuit drawn), it gives between 511 and 560 mA.

I've tested the circuit at 5 V and it is indeed fairly constant in current over that range of loads. I just can't figure out why my BJT is conducting without base current at 16 V. I can deal with the heat issue later, or upgrade BJTs if it can't be dealt with, but I think when it's only on for one second it should be fine.

I don't see why you have Q8 in the circuit. The simulation provided works without it.

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That's an LED driver to turn the circuit on and off.

OK, then try an NPN transistor instead of the PNP!

Q8 must be PNP because the LED driver sinks current but I think you're referring to Q7. I'm not sure how to make it constant current with an NPN.

Oh... My current NPN design depends on the current through Q8. Dangit.

I did mean try an NPN at Q8, but I didn't realize that the on/off signal was sent by another piece of equipment or circuit. I think the problem is that the on/off signal has to shift between 0V and about 15V so the Q8 PNP will switch properly. So my suggestion doesn't seem to be a good one. Sorry.